Try to Convey for Palindrome! ! ! ! !

given a string, characters can be shuffled to make a palindrome. !
What is the minimum possible number of insertions to original string needed so that it will be a palindrome (after shuffling, if required).

Input

T -> number of test cases
T number of Strings in different lines

import java.util.Arrays;
import java.io.InputStreamReader;
import java.io.BufferedReader;

public class Xsquare{
 public static void main (String[] args) throws Exception{
   BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  int limit = Integer.parseInt(br.readLine());
  
  int [] alphabets = new int[26];
  
  while(limit-- >;0){
   String input = br.readLine();
   Arrays.fill(alphabets,0);
   char [] inpChar = input.toCharArray();
   int sum = 0;
   
   for (int i=0;i<input.length();i++){
     int pos = (int)inpChar[i] - (int)'a';
     alphabets[pos]+=1;
   }
   
   for(int i=0;i<;26;i++){
    if(alphabets[i]%2==0)
     sum+=0;
    else
     sum+=1; 
   }
   
   if(sum<=0)
    sum=0;
   else 
    sum-=1;
   System.out.println(sum);
  }
 }
}

MaxiMizing XOR

Problem Statement

Given two integers, L and R, find the maximal values of A xor B, where Aand B satisfy the following condition:

L≤A≤B≤R

Input Format

The input contains two lines; L is present in the first line and R in the second line.

Constraints 
1≤L≤R≤103

Output Format

The maximal value as mentioned in the problem statement.

Sample Input

10
15

Sample Output

7

Explanation

The input tells us that L=10 and R=15. All the pairs which comply to above condition are the following: 
10⊕10=0 
10⊕11=1 
10⊕12=6 
10⊕13=7 
10⊕14=4 
10⊕15=5 
11⊕11=0 
11⊕12=7 
11⊕13=6 
11⊕14=5 
11⊕15=4 
12⊕12=0 
12⊕13=1 
12⊕14=2 
12⊕15=3 
13⊕13=0 
13⊕14=3 
13⊕15=2 
14⊕14=0 
14⊕15=1 
15⊕15=0 
Here two pairs (10, 13) and (11, 12) have maximum xor value 7, and this is the answer.

/*code written by Sudhakar Pandey */

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

#include <assert.h>

int maxXor(int l, int r) {

    int i,j,max=0,p=0;

    for(i=l;i<=r;i++)

        {

         for(j=i;j<=r;j++)

             {

               p=i^j;

               if(p>max)

                 max=p;  

         }

    }

 return max;

}

int main() {

    int res;

    int l;

    scanf("%d", &l);

    

    int r;

    scanf("%d", &r);

    

    res = maxXor(l, r);

    printf("%d", res);

    

    return 0;

}